Question: Evaluate the improper integral if it exists. $\int^{\infty}_{1}\dfrac1x\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $e$ (Choice D) D The improper integral diverges.
Answer: First, let's rewrite the improper integral: $\int^{\infty}_{1}\dfrac1x\,dx=\lim_{b\to\infty}\int_{1}^{b}\dfrac1x\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{2}^{\infty}\dfrac1{\sqrt x}\,dx}&=\lim_{b\to\infty}\int_{1}^{b}\dfrac1x\,dx\\ \\ \\ &=\lim_{b\to\infty}\Big[\ln(x)\Big]_{1}^b\\ \\ \\ &=\lim_{b\to\infty}\left(\ln(b)-\ln(1)\right)\\ \\ &=\lim_{b\to\infty}\left(\ln(b)\right)\\ \\ &=\infty \end{aligned}$ The answer: The improper integral diverges.